Mean-value theorem

[Kun Guo]

I do not quite understand step b and c(see attached) in the proof of Mean-value theorem. Did you use Green's identity or some other identity when dragging out terms from the integral?

[Victor Ivrii]

b. Gauss identity (you may call it Green in 2D). We proved this before.

c. $\Sigma$ is a sphere of radius $r$ with a center $y$ so $G$ so for $x\in \Sigma$  $G(x,y)=c |x-y|^{2-n}=cr^{2-n}$ is constant and could be moved out of integral.

[Ziting Zhou]

Hi Professor Ivrii. I have a question on the attachment above. In part b, after dragging out the factor, why the remainder is (partial u)/(partial v) instead of u? Thanks.

[Victor Ivrii]

Hi Professor Ivrii. I have a question on the attachment above. In part b, after dragging out the factor, why the remainder is (partial u)/(partial v) instead of u? Thanks.

Right, fixed and details added
http://www.math.toronto.edu/courses/apm346h1/20129/L26.html#sect-26.4

[Kun Guo]

thanks professor, now it makes more sense 

[Ziting Zhou]

thanks professor, but the formulae are not visible for me.

[Victor Ivrii]

Try again

[Ziting Zhou]

Thanks. It works now. 

Try again

[Thomas Nutz]

Hi,

what is meant by $\frac{\partial G}{\partial v_x}$ (eq. 7 in Lec. 26)? is $\frac{\partial G}{\partial v_x}=\nabla G \cdot (\nabla \cdot \vec{v})$?

Thanks!

[Thomas Nutz]

or rather $\nabla G \cdot  \frac{\partial}{\partial x}(\vec{v})$?

[Victor Ivrii]

or rather $\nabla G \cdot  \frac{\partial}{\partial x}(\vec{v})$?

No, $\nu_x$ means only that we differentiate with respect to $x$

[Tse Chiang Chen]

Hi Professor (and fellow classmates),
I am wondering the same question as Thomas; specifically, does it mean this? (See attached)

[Victor Ivrii]

Hi Professor (and fellow classmates),
I am wondering the same question as Thomas; specifically, does it mean this? (See attached)

No, what you are writing is wrong. Function $G(x,y)$ depends on both $x$ and $y$ and we need to differentiate with respect to $x$ not $y$. So in fact I mean is
$$
\nabla_x G(x,y) \cdot \nu(x)
$$
where $\nabla_x$ means gradient with respect to $x$. Here $y$ is considered as a parameter

[Thomas Nutz]

I see, thanks.

Are we considering closed domains, i.e. $\Sigma \in \Omega$? I think we have to, otherwise the $max_{\Omega}u \geq max_{\Sigma}u$ does not pull, correct?

[Victor Ivrii]

I see, thanks.

Are we considering closed domains, i.e. $\Sigma \in \Omega$? I think we have to, otherwise the $max_{\Omega}u \geq max_{\Sigma}u$ does not pull, correct?

Yes, it is exactly correct except $\Sigma \subset \Omega$. Alternatively we can write $\max_{\bar{\Omega}}u \geq \max_{\Sigma}u$ where $\bar{\Omega}=\Omega \cup \Sigma$.